Consider the fission of ${}^{235}$U induced by a thermal neutron: ${}^{235}\text{U}{+}^1\text{n}{\to }^{92}\text{Kr}{+}^{141}\text{Ba}+3{}^1\text{n}$ Atomic masses (in u): $m{(}^{235}\text{U})=235.0439$, $m{(}^1\text{n})=1.008665$, $m{(}^{92}\text{Kr})=91.9262$, $m{(}^{141}\text{Ba})=140.9144$.

Step-by-step solution

1. Write the total mass of reactants: $m_{\text{react}}=m{(}^{235}\text{U})+m{(}^1\text{n})=235.0439+1.008665=236.052565\text{u}$.
2. Write the total mass of products: $m_{\text{prod}}=m{(}^{92}\text{Kr})+m{(}^{141}\text{Ba})+3m{(}^1\text{n})=91.9262+140.9144+3\times 1.008665=236.(calculate)91.9262+140.9144=232.8406;3\times 1.008665=3.025995;sum=235.866595\text{u}$.
3. Find the mass defect: $\Delta m=m_{\text{react}}-m_{\text{prod}}=236.052565-235.866595=0.185970\text{u}$.
4. Convert mass defect to energy using $1\text{u}=931.5\text{MeV}/c^2$: $Q=0.185970\times 931.5=173.3\text{MeV}$. (Rounded to two significant figures, $\approx 200\text{MeV}$ is commonly quoted for ${}^{235}$U fission.)
5. For part (b), number of ${}^{235}$U atoms in 1.0 g: $N=\frac{1.0\text{g}}{235{\text{g mol}}^{-1}}\times N_A=\frac{1}{235}\times 6.022\times 10^{23}=2.56\times 10^{21}$ atoms.
6. Total energy released: $E=N\times Q\times 1.602\times 10^{-13}\text{J/MeV}$ (since $1\text{MeV}=1.602\times 10^{-13}\text{J}$). Using $Q\approx 200\text{MeV}$, $E=2.56\times 10^{21}\times 200\times 1.602\times 10^{-13}=8.2\times 10^{13}\text{J}$.
7. For part (c), power $P=1\text{GW}=1\times 10^9{\text{J s}}^{-1}$. Energy per fission $=200\text{MeV}=200\times 1.602\times 10^{-13}=3.20\times 10^{-11}\text{J}$.
8. Number of fissions per second $=P/E_{\text{fission}}=1\times 10^9/3.20\times 10^{-11}=3.1\times 10^{19}{\text{s}}^{-1}$.

Concept tested: Energy released in nuclear fission and power calculations

Published 11 September 2026 · How this solution was prepared and checked